IMO 2023 Solution Notes
Evan Chen《陳誼廷》
28 May 2024
This is a compilation of solutions for the 2023 IMO. The ideas of the
solution are a mix of my own work, the solutions provided by the competition
organizers, and solutions found by the community. However, all the writing
is maintained by me.
These notes will tend to be a bit more advanced and terse than the “official”
solutions from the organizers. In particular, if a theorem or technique is not
known to beginners but is still considered “standard”, then I often prefer to
use this theory anyways, rather than try to work around or conceal it. For
example, in geometry problems I typically use directed angles without further
comment, rather than awkwardly work around configuration issues. Similarly,
sentences like “let R denote the set of real numbers” are typically omitted
entirely.
Corrections and comments are welcome!
Contents
0 Problems
2
1 Solutions to Day 1
1.1 IMO 2023/1, proposed by Santiago Rodriguez (COL) . . . . . . . . . . . .
1.2 IMO 2023/2, proposed by Tiago Mourão and Nuno Arala (POR) . . . . .
1.3 IMO 2023/3, proposed by Ivan Chan (MAS) . . . . . . . . . . . . . . . . .
4
4
5
7
2 Solutions to Day 2
9
2.1 IMO 2023/4, proposed by Merlijn Staps (NLD) . . . . . . . . . . . . . . . 9
2.2 IMO 2023/5, proposed by Merlijn Staps and Daniël Kroes (NLD) . . . . . 11
2.3 IMO 2023/6, proposed by Ankan Bhattacharya, Luke Robitaille (USA) . 14
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§0 Problems
1. Determine all composite integers n > 1 that satisfy the following property: if
d1 < d2 < · · · < dk are all the positive divisors of n with then di divides di+1 + di+2
for every 1 ≤ i ≤ k − 2.
2. Let ABC be an acute-angled triangle with AB < AC. Let Ω be the circumcircle of
ABC. Let S be the midpoint of the arc CB of Ω containing A. The perpendicular
from A to BC meets BS at D and meets Ω again at E 6= A. The line through D
parallel to BC meets line BE at L. Denote the circumcircle of triangle BDL by ω.
Let ω meet Ω again at P 6= B. Prove that the line tangent to ω at P meets line
BS on the internal angle bisector of ∠BAC.
3. For each integer k ≥ 2, determine all infinite sequences of positive integers a1 , a2 ,
. . . for which there exists a polynomial P of the form
P (x) = xk + ck−1 xk−1 + · · · + c1 x + c0 ,
where c0 , c1 , . . . , ck−1 are non-negative integers, such that
P (an ) = an+1 an+2 · · · an+k
for every integer n ≥ 1.
4. Let x1 , x2 , . . . , x2023 be pairwise different positive real numbers such that
s
1
1
1
an = (x1 + x2 + · · · + xn )
+
+ ··· +
x1 x2
xn
is an integer for every n = 1, 2, . . . , 2023. Prove that a2023 ≥ 3034.
5. Let n be a positive integer. A Japanese triangle consists of 1 + 2 + · · · + n circles
arranged in an equilateral triangular shape such that for each 1 ≤ i ≤ n, the ith
row contains exactly i circles, exactly one of which is colored red. A ninja path in
a Japanese triangle is a sequence of n circles obtained by starting in the top row,
then repeatedly going from a circle to one of the two circles immediately below it
and finishing in the bottom row. Here is an example of a Japanese triangle with
n = 6, along with a ninja path in that triangle containing two red circles.
n=6
In terms of n, find the greatest k such that in each Japanese triangle there is a
ninja path containing at least k red circles.
6. Let ABC be an equilateral triangle. Let A1 , B1 , C1 be interior points of ABC
such that BA1 = A1 C, CB1 = B1 A, AC1 = C1 B, and
∠BA1 C + ∠CB1 A + ∠AC1 B = 480◦ .
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Let A2 = BC1 ∩ CB1 , B2 = CA1 ∩ AC1 , C2 = AB1 ∩ BA1 . Prove that if triangle
A1 B1 C1 is scalene, then the circumcircles of triangles AA1 A2 , BB1 B2 , and CC1 C2
all pass through two common points.
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IMO 2023 Solution Notes
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§1 Solutions to Day 1
§1.1 IMO 2023/1, proposed by Santiago Rodriguez (COL)
Available online at https://aops.com/community/p28097575.
Problem statement
Determine all composite integers n > 1 that satisfy the following property: if
d1 < d2 < · · · < dk are all the positive divisors of n with then di divides di+1 + di+2
for every 1 ≤ i ≤ k − 2.
The answer is prime powers.
¶ Verification that these work. When n = pe , we get di = pi−1 . The ith relationship
reads
pi−1 | pi + pi+1
which is obviously true.
¶ Proof that these are the only answers. Conversely, suppose n has at least two
distinct prime divisors. Let p < q denote the two smallest ones, and let pe be the largest
power of p which both divides n and is less than q, hence e ≥ 1. Then the smallest
factors of n are 1, p, . . . , pe , q. So we are supposed to have
n n
n
(p + 1)n
| e + e−1 =
q p
p
pe
which means that the ratio
q(p + 1)
pe
needs to be an integer, which is obviously not possible.
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§1.2 IMO 2023/2, proposed by Tiago Mourão and Nuno Arala (POR)
Available online at https://aops.com/community/p28097552.
Problem statement
Let ABC be an acute-angled triangle with AB < AC. Let Ω be the circumcircle of
ABC. Let S be the midpoint of the arc CB of Ω containing A. The perpendicular
from A to BC meets BS at D and meets Ω again at E 6= A. The line through D
parallel to BC meets line BE at L. Denote the circumcircle of triangle BDL by ω.
Let ω meet Ω again at P 6= B. Prove that the line tangent to ω at P meets line BS
on the internal angle bisector of ∠BAC.
Claim — We have LP S collinear.
Proof. Because ]LP B = ]LDB = ]CBD = ]CBS = ]SCB = ]SP B.
Let F be the antipode of A, so AM F S is a rectangle.
Claim — We have P DF collinear. (This lets us erase L.)
Proof. Because ]SP D = ]LP D = ]LBD = ]SBE = ]F CS = ]F P S.
Let us define X = AM ∩ BS and complete chord P XQ. We aim to show that P XQ
is tangent to (P DLB).
S
A
P
X
Q
D
L
B
C
E
F
M
Claim (Main projective claim) — We have XP = XA.
Proof. Introduce Y = P DF ∩ AM . Note that
A
F
−1 = (SM ; EF ) = (S, X; D, AF ∩ ES) = (∞X; Y A)
where ∞ = AM ∩ SF is at infinity (because AM SF is a rectangle). Thus, XY = XA.
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S
A
P
X
D
Y
E
F
M
Since 4AP Y is also right, we get XP = XA.
Alternative proof of claim without harmonic bundles, from Solution 9 of the marking scheme.
With Y = P DF ∩ AM defined as before, note that AE k SM and AM k SF (as AM F S
is a rectangle) gives respectively the similar triangles
4AXD ∼ 4M XS,
4XDY ∼ 4SDF.
From this we conclude
AX
AX + XM
AM
SF
XY
=
=
=
=
.
XD
XD + SX
SD
SD
XD
So AX = XY and as before we conclude XP = XA.
¯
˜ have the same measure. Since AS
ˆ
From XP = XA, we conclude that P
M and AQ
¯ have the same measure, it follows P
˜
˜ have the same measure. The
and EM
E and SQ
desired tangency then follows from
]QP L = ]QP S = ]P QE = ]P F E = ]P DL.
Remark (Logical ordering). This solution is split into two phases: the “synthetic phase”
where we do a bunch of angle chasing, and the “projective phase” where we use cross-ratios
because I like projective. For logical readability (so we write in only one logical direction),
the projective phase is squeezed in two halves of the synthetic phase, but during an actual
solve it’s expected to complete the whole synthetic phase first (i.e. to reduce the problem to
show XP = XA).
Remark. There are quite a multitude of approaches for this problem; the marking scheme
for this problem at the actual IMO had 13 different solutions.
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§1.3 IMO 2023/3, proposed by Ivan Chan (MAS)
Available online at https://aops.com/community/p28097600.
Problem statement
For each integer k ≥ 2, determine all infinite sequences of positive integers a1 , a2 ,
. . . for which there exists a polynomial P of the form
P (x) = xk + ck−1 xk−1 + · · · + c1 x + c0 ,
where c0 , c1 , . . . , ck−1 are non-negative integers, such that
P (an ) = an+1 an+2 · · · an+k
for every integer n ≥ 1.
The answer is an being an arithmetic progression. Indeed, if an = d(n − 1) + a1 for d ≥ 0
and n ≥ 1, then
an+1 an+2 . . . an+k = (an + d)(an + 2d) . . . (an + kd)
so we can just take P (x) = (x + d)(x + 2d) . . . (x + kd).
The converse direction takes a few parts.
Claim — Either a1 < a2 < · · · or the sequence is constant.
Proof. Note that
P (an−1 ) = an an+1 · · · an+k−1
P (an ) = an+1 an+2 · · · an+k
P (an )
=⇒ an+k =
· an .
P (an−1 )
Now the polynomial P is strictly increasing over N.
So assume for contradiction there’s an index n such that an < an−1 . Then in fact the
above equation shows an+k < an < an−1 . Then there’s an index ` ∈ [n + 1, n + k] such
that a` < a`−1 , and also a` < an . Continuing in this way, we can an infinite descending
subsequence of (an ), but that’s impossible because we assumed integers.
Hence we have a1 ≤ a2 ≤ · · · . Now similarly, if an = an−1 for any index n, then
an+k = an , ergo an−1 = an = an+1 = · · · = an+k . So the sequence is eventually constant,
and then by downwards induction, it is fully constant.
Claim — There exists a constant C (depending only P , k) such that we have
an+1 ≤ an + C.
Proof. Let C be a constant such that P (x) < xk + Cxk−1 for all x ∈ N (for example
C = c0 + c1 + · · · + ck−1 + 1 works). We have
an+k =
P (an )
an+1 an+2 . . . an+k−1
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P (an )
(an + 1)(an + 2) . . . (an + k − 1)
akn + C · ak−1
n
<
(an + 1)(an + 2) . . . (an + k − 1)
< an + C + 1.
<
Assume henceforth an is nonconstant, and hence unbounded. For each index n and term
an in the sequence, consider the associated differences d1 = an+1 − an , d2 = an+2 − an+1 ,
. . . , dk = an+k − an+k−1 , which we denote by
∆(n) := (d1 , . . . , dk ).
This ∆ can only take up to C k different values. So in particular, some tuple (d1 , . . . , dn )
must appear infinitely often as ∆(n); for that tuple, we obtain
P (aN ) = (aN + d1 )(aN + d1 + d2 ) . . . (aN + d1 + · · · + dk )
for infinitely many N . But because of that, we actually must have
P (X) = (X + d1 )(X + d1 + d2 ) . . . (X + d1 + · · · + dk ).
However, this also means that exactly one output to ∆ occurs infinitely often (because
that output is determined by P ). Consequently, it follows that ∆ is eventually constant.
For this to happen, an must eventually coincide with an arithmetic progression of some
common difference d, and P (X) = (X + d)(X + 2d) . . . (X + kd). Finally, this implies by
downwards induction that an is an arithmetic progression on all inputs.
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§2 Solutions to Day 2
§2.1 IMO 2023/4, proposed by Merlijn Staps (NLD)
Available online at https://aops.com/community/p28104298.
Problem statement
Let x1 , x2 , . . . , x2023 be pairwise different positive real numbers such that
s
1
1
1
an = (x1 + x2 + · · · + xn )
+
+ ··· +
x1 x2
xn
is an integer for every n = 1, 2, . . . , 2023. Prove that a2023 ≥ 3034.
qP
Pn 1
n
Note that an+1 >
1 xi
1 xi = an for all n, so that an+1 ≥ an + 1. Observe a1 = 1.
We are going to prove that
for all m ≥ 0
a2m+1 ≥ 3m + 1
by induction on m, with the base case being clear.
We now present two variations of the induction. The first shorter solution compares
an+2 directly to an , showing it increases by at least 3. Then we give a longer approach
that compares an+1 to an , and shows it cannot increase by 1 twice in a row.
¶ Induct-by-two solution.
Let u =
with three terms:
"
a2n+2 =
≥
q
xn+1
xn+2 6= 1. Note that by using Cauchy-Schwarz
#" (x1 + · · · + xn ) + xn+1 + xn+2
1
1
+ ··· +
x1
xn
s
r
(x1 + · · · + xn )
1
1
+ ··· +
x1
xn
r
+
xn+1
+
xn+2
+
1
xn+2
!2
+
1
#
xn+1
xn+2
xn+1
1 2
= an + u +
.
u
1
=⇒ an+2 ≥ an + u + > an + 2
u
where the last equality u + u1 > 2 is by AM-GM, strict as u 6= 1. It follows that
an+2 ≥ an + 3, completing the proof.
¶ Induct-by-one solution.
The main claim is:
Claim — It’s impossible to have an = c, an+1 = c + 1, an+2 = c + 2 for any c and n.
Proof. Let p = xn+1 and q = xn+2 for brevity. Let s =
c2 = a2n = st.
From an = c and an+1 = c + 1 we have
1
(c + 1)2 = a2n+1 = (p + s)
+t
p
9
1 xi and t =
Pn
1 xn , so
Pn 1
IMO 2023 Solution Notes
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1
1
= st + pt + s + 1 = c2 + pt + s + 1
p
p
√
√
AM-GM
≥ c2 + 2 st + 1 = c2 + 2 c2 + 1 = (c + 1)2 .
Hence, equality must hold in the AM-GM we must have exactly
pt =
1
s = c.
p
If we repeat the argument again on an+1 = c + 1 and an+2 = c + 2, then
1
1
q
+ t = (p + s) = c + 1.
p
q
However this forces pq = pq = 1 which is impossible.
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§2.2 IMO 2023/5, proposed by Merlijn Staps and Daniël Kroes (NLD)
Available online at https://aops.com/community/p28104367.
Problem statement
Let n be a positive integer. A Japanese triangle consists of 1 + 2 + · · · + n circles
arranged in an equilateral triangular shape such that for each 1 ≤ i ≤ n, the ith
row contains exactly i circles, exactly one of which is colored red. A ninja path in a
Japanese triangle is a sequence of n circles obtained by starting in the top row, then
repeatedly going from a circle to one of the two circles immediately below it and
finishing in the bottom row. Here is an example of a Japanese triangle with n = 6,
along with a ninja path in that triangle containing two red circles.
n=6
In terms of n, find the greatest k such that in each Japanese triangle there is a ninja
path containing at least k red circles.
The answer is
k = blog2 (n)c + 1.
¶ Construction. It suffices to find a Japanese triangle for n = 2e − 1 with the property
that at most e red circles in any ninja path. The construction shown below for e = 4
obviously generalizes, and works because in each of the sets {1}, {2, 3}, {4, 5, 6, 7}, . . . ,
{2e−1 , . . . , 2e − 1}, at most one red circle can be taken. (These sets are colored in different
shades of red for visual clarity).
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¶ Bound. Conversely, we show that in any Japanese triangle, one can find a ninja path
containing at least
k = blog2 (n)c + 1.
The following short solution was posted at https://aops.com/community/p28134004,
apparently first found by the team leader for Iran.
We construct a rooted binary tree T1 on the set of all circles as follows. For each row,
other than the bottom row:
• Connect the red circle to both circles under it;
• White circles to the left of the red circle in its row are connected to the left;
• White circles to the right of the red circle in its row are connected to the right.
The circles in the bottom row are all leaves of this tree. For example, the n = 6
construction in the beginning gives the tree shown on the left half of the figure below:
T1
T2
Now focus on only the red circles, as shown in the right half of the figure. We build
a new rooted tree T2 where each red circle is joined to the red circle below it if there
was a path of (zero or more) white circles in T1 between them. Then each red circle has
at most 2 direct descendants in T2 . Hence the depth of the new tree T2 exceeds log2 (n),
which produces the desired path.
¶ Another recursive proof of bound, communicated by Helio Ng. We give another
proof that blog2 nc + 1 is always achievable. Define f (i, j) to be the maximum number
of red circles contained in the portion of a ninja path from (1, 1) to (i, j), including the
endpoints (1, 1) and (i, j). (If (i, j) is not a valid circle in the triangle, define f (i, j) = 0
for convenience.) An example is shown below with the values of f (i, j) drawn in the
circles.
1
2
2
3
3
3
3
2
2
3
3
3
1
2
3
3
3
2
2
2
3
3
12
2
3
3
2
3
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We have that
(
1 if (i, j) is red
f (i, j) = max {f (i − 1, j − 1), f (i, j − 1)} +
0 otherwise
since every ninja path passing through (i, j) also passes through either (i − 1, j − 1) or
(i, j − 1). Now consider the quantity Sj = f (0, j) + · · · + f (j, j). We obtain the following
recurrence:
l m
S
Claim — Sj+1 ≥ Sj + jj + 1.
Proof. Consider a maximal element f (m, j) of {f (0, j), . . . , f (j, j)}. We perform the
following manipulations:
(
j+1
j+1
X
X
1 if (i, j + 1) is red
Sj+1 =
max {f (i − 1, j), f (i, j)} +
0 otherwise
i=0
i=0
=
m
X
i=0
≥
m
X
i=0
j
X
max {f (i − 1, j), f (i, j)} +
max {f (i − 1, j), f (i, j)} + 1
i=m+1
f (i, j) +
j
X
f (i − 1, j) + 1
i=m+1
= Sj + f (m, j) + 1
Sj
≥ Sj +
+1
j
where the last inequality is due to Pigeonhole.
This is actually enough to solve the problem. Write n = 2c + r, where 0 ≤ r ≤
2c − 1.
Claim — Sn ≥ cn + 2r + 1. In particular,
S n
n
≥ c + 1.
Proof. First note that Sn ≥ cn + 2r + 1 implies Snn ≥ c + 1 because
Sn
cn + 2r + 1
2r + 1
≥
=c+
= c + 1.
n
n
n
We proceed by induction on n. The base case n = 1 is clearly true as S1 = 1. Assuming
that the claim holds for some n = j, we have
Sj
Sj+1 ≥ Sj +
+1
j
≥ cj + 2r + 1 + c + 1 + 1
= c(j + 1) + 2(r + 1) + 1
so the claim is proved for n = j + 1 if j + 1 is not a power of 2. If j + 1 = 2c+1 , then by
writing c(j + 1) + 2(r + 1) + 1 = c(j + 1) + (j + 1) + 1 = (c + 2)(j + 1) + 1, the claim is
also proved.
Now Snn ≥ c + 1 implies the existence of some ninja path containing at least c + 1
red circles, and we are done.
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§2.3 IMO 2023/6, proposed by Ankan Bhattacharya, Luke Robitaille (USA)
Available online at https://aops.com/community/p28104331.
Problem statement
Let ABC be an equilateral triangle. Let A1 , B1 , C1 be interior points of ABC such
that BA1 = A1 C, CB1 = B1 A, AC1 = C1 B, and
∠BA1 C + ∠CB1 A + ∠AC1 B = 480◦ .
Let A2 = BC1 ∩ CB1 , B2 = CA1 ∩ AC1 , C2 = AB1 ∩ BA1 . Prove that if triangle
A1 B1 C1 is scalene, then the circumcircles of triangles AA1 A2 , BB1 B2 , and CC1 C2
all pass through two common points.
This is the second official solution from the marking scheme, also communicated to me
by Michael Ren. Define O as the center of ABC and set the angles
α := ∠A1 CB = ∠CBA1
β := ∠ACB1 = ∠B1 AC
γ := ∠C1 AB = ∠C1 BA
so that
α + β + γ = 30◦ .
In particular, max(α, β, γ) < 30◦ , so it follows that A1 lies inside 4OBC, and similarly
for the others. This means for example that C1 lies between B and A2 , and so on.
Therefore the polygon A2 C1 B2 A1 C2 B1 is convex.
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A
β
γ
Y
A2
X
C1
B1
O
γ
B2
C2
β
A1
B
α
α
C
We start by providing the “interpretation” for the 480◦ angle in the statement:
Claim — Point A1 is the circumcenter of 4A2 BC, and similarly for the others.
Proof. We have ∠BA1 C = 180◦ − 2α, and
∠BA2 C = 180◦ − ∠CBC1 − ∠B1 CB
= 180◦ − (60◦ − γ) − (60◦ − β)
1
= 60◦ + β + γ = 90◦ − α = ∠BA1 C.
2
Since A1 lies inside 4BA2 C, it follows A1 is exactly the circumcenter.
Claim — Quadrilateral B2 C1 B1 C can be inscribed in a circle, say γa . Circles γb
and γc can be defined similarly. Finally, these three circles are pairwise distinct.
Proof. Using directed angles now, we have
]B2 B1 C2 = 180◦ − ]AB1 B2 = 180◦ − 2]ACB = 180◦ − 2(60◦ − α) = 60◦ + 2α.
By the same token, ]B2 C1 C2 = 60◦ + 2α. This establishes the existence of γa .
The proof for γb and γc is the same. Finally, to show the three circles are distinct, it
would be enough to verify that the convex hexagon A2 C1 B2 A1 C2 B1 is not cyclic.
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Assume for contradiction it was cyclic. Then
360◦ = ∠C2 A1 B1 + ∠B2 C1 A2 + ∠A2 B1 C2 = ∠BA1 C + ∠CB1 A + ∠AC1 B = 480◦
which is absurd. This contradiction eliminates the degenerate case, so the three circles
are distinct.
For the remainder of the solution, let Pow(P, ω) denote the power of a point P with
respect to a circle ω.
Let line AA1 meet γb and γc again at X and Y , and set ka := AX
AY . Consider the locus
of all points P such that
n
o
Ca := points P in the plane satisfying Pow(P, γb ) = ka Pow(P, γc ) .
We recall the coaxiality lemma 1 , which states that (given γb and γc are not concentric)
the locus Ca must be either a circle (if ka 6= 1) or a line (if ka = 1).
On the other hand, A1 , A2 , and A all obviously lie on Ca . (For A1 and A2 , the
powers are both zero, and for the point A, we have Pow(P, γb ) = AX · AA1 and
Pow(P, γc ) = AY · AA1 .) So Ca must be exactly the circumcircle of 4AA1 A2 from the
problem statement.
We turn to evaluating ka more carefully. First, note that
∠A1 XB1 = ∠A1 B2 B1 = ∠CB2 B1 = 90◦ − ∠B2 AC = 90◦ − (60◦ − γ) = 30◦ + γ.
Now using the law of sines, we derive
AX
sin ∠AB1 X
sin(∠A1 XB1 − ∠XAB1 )
=
=
AB1
sin ∠AXB1
sin ∠A1 XB1
sin ((30◦ + γ) − (30◦ − β))
sin(β + γ)
=
=
.
◦
sin(30 + γ)
sin(30◦ + γ)
sin(β+γ)
Similarly, AY = AC1 · sin(30
◦ +β) , so
ka =
AX
AB1 sin(30◦ + β)
=
·
.
AY
AC1 sin(30◦ + γ)
Now define analogous constants kb and kc and circles Cb and Cc . Owing to the symmetry
of the expressions, we have the key relation
ka kb kc = 1.
In summary, the three circles in the problem statement may be described as
Ca = (AA1 A2 ) = {points P such that Pow(P, γb ) = ka Pow(P, γc )}
Cb = (BB1 B2 ) = {points P such that Pow(P, γc ) = kb Pow(P, γa )}
Cc = (CC1 C2 ) = {points P such that Pow(P, γa ) = kc Pow(P, γb )} .
Since ka , kb , kc have product 1, it follows that any point on at least two of the circles must
lie on the third circle as well. The convexity of hexagon A2 C1 B2 A1 C2 B1 mentioned earlier
ensures these any two of these circles do intersect at two different points, completing the
solution.
1
We quickly outline a proof of this lemma: in the Cartesian coordinate system, the expression Pow((x, y), ω)
is an expression of the form x2 + y 2 + •x + •y + • for some constants • whose value does not matter.
b)
Substituting this into the equation ka Pow(P,γkac )−Pow(P,γ
= 0 gives the equation of a circle provided
−1
ka 6= 1, and when ka = 1, one instead recovers the radical axis.
16